SOLVING LOGARITHMS - MADE EASY WITH SOLVED EXAMPLES, EXERCISES, LINKS TO FURTHER STUDY

Please study

Logarithm Formulas before Solving Logarithms

if you have not already done so.

Having the knowledge of the Formulae
is a prerequisite here.

Example 1 of Solving Logarithms :

Express the following as single logarithms to any base.
(i) (1⁄2) log (a + b) - (1⁄2) log (a - b)
(ii) log (a + b) + log (a² - ab + b²) - log (x - y) - log (x² + xy + y²)
(iii) (3⁄2) log x - (1⁄3) log y + (2⁄3) log z - (1⁄5) log a

Solution to Example 1 of Solving Logarithms :

Solution to 1(i) of Solving Logarithms :

Let P = (1⁄2) log (a + b) - (1⁄2) log (a - b)
We know in log of a power (see Formula 7) , the exponent multipliesthe log. Considering the reverse, If a quantity multiplies thelog, it becomes exponent.
∴ P = log {(a + b)^(1⁄2)} - log {(a - b)^(1⁄2)} We know, log of a quotient can be written as the difference of the logs of the numerator and denominator of the quotient (See Formula 6)
Considering the reverse, the difference of the logs of different quantitiescan be written as log of a quotient (the quantity whose log has negative sign, in the denominator) of the quantities.
= log {(a + b)^(1⁄2)}⁄{(a - b)^(1⁄2)}
= log {(a + b)⁄(a - b)}^(1⁄2). Ans.

Solution to 1(ii) of Solving Logarithms :

Let P = log (a + b) + log (a² - ab + b²) - log (x - y) - log (x² + xy + y²)
= log (a + b) + log (a² - ab + b²) - {log (x - y) + log (x² + xy + y²)}
We know, log of a product can be written as the sum of the logs of the factors of the product (See Formula 5)
Considering the reverse, the sum of the logs of different quantitiescan be written as log of product of the quantities.
∴ P = log {(a + b)(a² - ab + b²)} - log {(x - y)(x² + xy + y²)}
We know from Polynomials, (a + b)(a² - ab + b²) = a³ + b³;
and (x - y)(x² + xy + y²) = x³ - y³
∴ P = log (a³ + b³) - log (x³ - y³)
We know, log of a quotient can be written as the difference of the logs of the numerator and denominator of the quotient (See Formula 6)
Considering the reverse, the difference of the logs of different quantitiescan be written as log of a quotient (the quantity whose log has negative sign, in the denominator) of the quantities.
∴ P= log {(a³ + b³)⁄(x³ - y³)} Ans.

Solution to 1(iii) of Solving Logarithms :

Let P = (3⁄2) log x - (1⁄3) log y + (2⁄3) log z - (1⁄5) log a
We know in log of a power (see Formula 7) , the exponent multipliesthe log. Considering the reverse, If a quantity multiplies thelog, it becomes exponent.
∴ P = log x^(3⁄2) + log z^(2⁄3) - {log y^(1⁄3) + log a^(1⁄5)}
We know, log of a product can be written as the sum of the logs of the factors of the product (See Formula 5)
Considering the reverse, the sum of the logs of different quantitiescan be written as log of product of the quantities.
∴ P = log {x^(3⁄2) x z^(2⁄3)} - log {y^(1⁄3) x a^(1⁄5)}
We know, log of a quotient can be written as the difference of the logs of the numerator and denominator of the quotient (See Formula 6)
Considering the reverse, the difference of the logs of different quantitiescan be written as log of a quotient (the quantity whose log has negative sign, in the denominator) of the quantities.
∴ P = log [{x^(3⁄2) x z^(2⁄3)}⁄{y^(1⁄3) x a^(1⁄5)}] Ans.

Great Deals on School & Homeschool Curriculum Books

Get The Best Grades With the Least Amount of Effort

Here is a collection of proven tips,
tools and techniques to turn you into
a super-achiever - even if you've never
thought of yourself as a "gifted" student.

The secrets will help you absorb, digest
and remember large chunks of information
quickly and easily so you get the best grades
with the least amount of effort.

If you apply what you read from the above
collection, you can achieve best grades without
giving up your fun, such as TV, surfing the net,
playing video games or going out with friends!

Know more about the Speed Study System.

Example 2 of Solving Logarithms :

Do the following problems on Solving Logarithms :

Solve
(i) 3^{(2x + 1)}.4^{(4x - 1)} = 36
(ii) 4^{(log₉ 3)} + 9^{(log₂ 4)} = 10^{(log_x 83)}

Solution to Example 2 of Solving Logarithms :

Solution to 2(i) of Solving Logarithms :

3^{(2x + 1)}.4^{(4x - 1)} = 36
Taking logarithms on both sides, we get
log {3^{(2x + 1)}.4^{(4x - 1)}} = log 36
⇒ log {3^{(2x + 1)}} + log {4^{(4x - 1)}} = log 36
We know in log of a power (see Formula 7) , the exponent multipliesthe log.
∴ (2x + 1) log 3 + (4x - 1) log 4 = log 36
⇒ 2x log 3 + 4x log 4 + log 3 - log 4 = log 36
⇒ 2x(log 3 + 2 log 4) = log 36 - log 3 + log 4
We know in log of a power (see Formula 7) , the exponent multipliesthe log. Considering the reverse, If a quantity multiplies thelog, it becomes exponent.
∴ L.H.S. = 2x(log 3 + log 4²)
We know, log of a product can be written as the sum of the logs of the factors of the product (See Formula 5)
Considering the reverse, the sum of the logs of different quantitiescan be written as log of product of the quantities.
We know, log of a quotient can be written as the difference of the logs of the numerator and denominator of the quotient (See Formula 6)
Considering the reverse, the difference of the logs of different quantitiescan be written as log of a quotient (the quantity whose log has negative sign, in the denominator) of the quantities.
Applying these to L.H.S. and R.H.S., we get
2x{log (3 x 4²)} = log (36 x 4⁄3) = log 48
⇒ 2x{log 48} = log 48 ⇒ 2x = 1 ⇒ x = 1⁄2. Ans.

Alternative Solution: This problem can be solved from Exponents knowledge without using logarithms as given below.
3^{(2x + 1)}.4^{(4x - 1)} = 36 ⇒ 3^(2x).3¹.4^(4x).4^(-1) = 36
⇒ (3)^2x.(4²)^2x.(3⁄4) = 36
⇒ 3^2x.(16)^2x = 36 x (4⁄3) = 48 ⇒ (3 x 16)^2x = 48
⇒ (48)^2x = 48 ⇒ 2x = 1 ⇒ x = 1⁄2. Ans.

Solution to 2(ii) of Solving Logarithms :

The given equation is 4^{(log₉ 3)} + 9^{(log₂ 4)} = 10^{(log_x 83)}
We have log₉ 3 = log_3² 3¹
We know, in log of a power to the base of another power (see Formula 9),the quotient of the exponents (exponent of the base in the denominator) multiplies the log.
∴ log₉ 3 = (1⁄2) log₃ 3
We know log of a quantity to the same base is 1. (See Formula 4)
∴ log₉ 3 = (1⁄2) (1) = 1⁄2;
log₂ 4 = log₂ 2²
We know in log of a power (see Formula 7) , the exponent multipliesthe log.
∴ log₂ 4 = 2 log₂ 2
We know log of a quantity to the same base is 1. (See Formula 4)
∴ log₂ 4 = 2 (1) = 2
∴ the given equation becomes 4^(1⁄2) + 9² = 10^{(log_x 83)}
⇒ 2 + 81 = 10^{(log_x 83)}
⇒ 83 = 10^{(log_x 83)} .......(i)
By converting the exponential form of (i) to logarithmic form (See Formula 1), we get
log_x 83 = log₁₀ 83
When two logarithms of same quantities are equal, their bases have to be equal.
∴ x = 10. Ans.
You can also apply Formula 2 to equation (i), to conclude x = 10.Ans.

More Solved Examples : Solving Logarithms

The following Link takes you
to more Solved Examples.

More Solved Examples

Research-based personalized
Math Help tutoring program :
Solving Logarithms

Here is a resource for Solid Foundation in
Math Fundamentals from Middle thru High School.
You can check your self by the

FREE TRIAL.

Are you spending lot of money for math tutors to your
child and still not satisfied with his/her grades ?

Do you feel that more time from the tutor and
more personalized Math Help to identify and fix
the problems faced by your child will help ?

Here is a fool proof solution I strongly recommend
and that too With a minuscule fraction of the amount
you spent on tutors with unconditional 100% money
back Guarantee, if you are not satisfied.

SUBSCRIBE, TEST, IF NOT SATISFIED, RETURN FOR FULL REFUND

It is like having an unlimited time from an excellent Tutor.

It is an Internet-based math tutoring software program
that identifies exactly where your child needs help and
then creates a personal instruction plan tailored to your
child’s specific needs.

If your child can use a computer and access
the Internet, he or she can use the program.
And your child can access the program anytime
from any computer with Internet access.

Unique program to help improve math skills quickly and painlessly.

There is an exclusive, Parent Information Page provides YOU
with detailed reports of your child’s progress so you can
monitor your child’s success and give them encouragement.
These Reports include

• Time spent using the program
• Assessment results
• Personalized remediation curriculum designed for your child
• Details the areas of weakness where your child needs additional help
• Provides the REASONS WHY your child missed a concept
• List of modules accessed and amount of time spent in each module
• Quiz results
• Creates reports that can be printed and used to discuss issues with your child’s teachers

Personalized remediation curriculum designed for your child

Thus The features of this excellent Tutoring program are

• Using detailed testing techniques
• Identifing exactly where a student needs help.
• Its unique, smart system pinpointing precise problem areas -
• slowly and methodically guiding the student
• raising to the necessary levels to fix the problem.

Its research-based results have proven that
it really works for all students! in improving
math skills and a TWO LETTER GRADE INCREASE in
math test scores!,if they invest time in using
the program.

Proven for More than 10,000 U.S. public school
students who increased their math scores.

Proven methodology!

Exercise : Solving Logarithms

Solve the following problems on Solving Logarithms

1. Solve 4^x + 2^{(2x - 1)} = 3^{(x + 1⁄2)} + 3^{(x - 1⁄2)}
   (i) by using logarithms (ii) without using logarithms.
   
2. If (log_y x.log_z x - log_x x) + (log_z y.log_x y - log_y y) + (log_x z.log_y z - log_z z) = 0,
   prove that xyz = 1.

For Answer to problem1 of the above exercise on
Solving Logarithms see at the bottom of the page.

Progressive Learning of Math : Solving Logarithms

Recently, I have found a series of math curricula
(Both Hard Copy and Digital Copy) developed by a Lady Teacher
who taught everyone from Pre-K students to doctoral students
and who is a Ph.D. in Mathematics Education.

This series is very different and advantageous
over many of the traditional books available.
These give students tools that other books do not.
Other books just give practice.
These teach students “tricks” and new ways to think.

These build a student’s new knowledge of concepts
from their existing knowledge.
These provide many pages of practice that gradually
increases in difficulty and provide constant review.

These also provide teachers and parents with lessons
on how to work with the child on the concepts.

The series is low to reasonably priced and include

Elementary Math curriculum

and

Algebra Curriculum.

Answers to Exercise : Solving Logarithms

Answer to problem on Solving Logarithms :

(1) 3⁄2